6
$\begingroup$

Given a regular tessellation, i.e. either a platonic solid (a tessellation of the sphere), the tessellation of the euclidean plane by squares or by regular hexagons, or a regular tessellation of the hyperbolic plane.

One can consider its isometry group $G$. It acts on the set of all faces $F$. I want to define a symmetric coloring of the tessellation as a surjective map from $c:F\rightarrow C$ to a finite set of colors $C$, such that for each group element $G$ there is a permutation $p_g$ of the colors, such that $c(gx)=p_g\circ c(x)$. ($p:G\rightarrow $Sym$(C)$ is a group homomorphism).

Examples for such colorings are the trivial coloring $c:F\rightarrow \{1\}$ or the coloring of the plane as an infinite chessboard. The only nontrivial symmetric colorings of the tetrahedron, is the one, that assigns a different color to each face. For the other platonic solids there are also those colorings that assign the same colors only to opposite faces.

So my question is: Does every regular tessellation of the hyperbolic plane admit a nontrivial symmetric coloring?

I wanted to write a computer program that visualizes those tessellations, but I didn't find a good strategy which colors should be used. So I came up with this question.

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
5
$\begingroup$

The answer is yes. Moreover, for every two different faces $A$ and $B$ there is a symmetric coloring assigning different colors to $A$ and $B$.

The isometry group $G$ is residually finite, hence here is a normal finite index subgroup $H$ of $G$ that contains no elements (except the identity) sending $A$ to itself or to $B$. Assign a unique color to each orbit of $H$.

The coloring symmetry condition is essentially he following: if $f\in G$ and faces $X$ and $Y$ are of the same color, then so are $f(X)$ and $f(Y)$. Since $X$ and $Y$ are of the same color, there exists $h\in H$ such that $h(X)=Y$. Since $H$ is normal, $h_1:=f^{-1}hf\in H$. But $h_1(F(X))=F(Y)$, hence $F(X)$ and $F(Y)$ are of the same color.

Improve this answer
$\endgroup$
6
  • $\begingroup$ as Daniel mentioned below: The faces can be identified with $G/$Stab$(x)$, for any face $x$. Coloring all faces of $G/H$ in the same color is only senseful, if $H$ contains $Stab(x)$. So one should ask, what the finite index subgroups of $G$, that contain $H$ are. eg. the subgroup of the symmetries of the 3-simplex (0,1,2,3) generated by the transposition (0,1) does not give such a coloring. $\endgroup$
    – HenrikRüping
    Commented Jun 23, 2010 at 16:55
  • $\begingroup$ I think normality tells you, that any isometry, that leaves one color unchanged already leaves all colors unchanged, e.g. in the tetrahedron case $H=Stab(3)\cong S_3$ gives a senseful coloring, although $S_3$ is not normal in $S^4$. $\endgroup$
    – HenrikRüping
    Commented Jun 23, 2010 at 16:56
  • $\begingroup$ No, the construction is different from Daniel's, and here you don't need $H$ to contain any stabilizer. Consider the following example on the plane tiled by squares: $H$ is the subgroup consisting of all parallel translations by vectors of the form $(a,b)$ where $a+b$ is even. There are two orbits, corresponding to black and white colors in the standard checkerboard coloring. The subgroup acts simply transitively on each orbit. $\endgroup$
    – Sergei Ivanov
    Commented Jun 23, 2010 at 17:07
  • $\begingroup$ I assume in this example you let $G$ be the group of all translations, which is not the full symmetry group of the tilings, which is $\mathbb{Z}^2\rtimes D_8$. So if you consider the subgroup generated by translations of $(3,0)$ and $(0,2)$. You get 6 cosets (=colors) colors . But a rotation of 90° around the origin doesn't map squares of one color to squares of another color, as (0,0) and (3,0) lie in the same coset, while (0,0) and (0,3) don't. $\endgroup$
    – HenrikRüping
    Commented Jun 23, 2010 at 21:33
  • $\begingroup$ The group generated by (3,0) and (0,2) is not normal in the full symmetry group. (Because the conjugation by the rotation does not preserve it.) This is why it does not work. The group in my example is normal in the full symmetry group. $\endgroup$
    – Sergei Ivanov
    Commented Jun 23, 2010 at 22:16
1
$\begingroup$

As the tesselation is regular, its symmetry group G acts transitively; let H be a subgroup strictly containing the stabilizer of a face. Then the orbit of the stabilized face under H, and its translates by other elements of the group, form a symmetric coloring---if H is of finite index and does not act transitively, it gives a coloring of the form you want. In the case of a tetrahedron, any group strictly containing the stabilizer of a face acts transitively on the faces.

More conceptually, in any case but a tetrahedron, quotienting by the stabilizer gives a group that acts simply transitively on the faces, and so is isomorphic to the set of faces---the colored sets are just the cosets.

So the question becomes---does such a subgroup exist in the triangle groups (the group of symmetries of a regular tesselation of the hyperbolic plane)?

Improve this answer
$\endgroup$
1
  • $\begingroup$ Ah, I was unclear. Edited. $\endgroup$
    – Daniel Litt
    Commented Jun 22, 2010 at 23:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .